9 min readConcept sheetA 3-mark numerical or part of a 5-mark question; power and the dioptre come as 1-mark and 2-mark questions

Lens formula, magnification and power of a lens

For lenses use the lens formula 1/v − 1/u = 1/f and m = v/u. Power P = 1/f, with f in metres, is measured in dioptres (D).

Lens formula1/v − 1/u = 1/f

Magnificationm = h′/h = v/u

PowerP = 1/f (f in metres); unit dioptre, 1 D = 1 m⁻¹

SignsConvex: f and P positive; concave: f and P negative

Mirror formula vs lens formula

Mirror1/v + 1/u = 1/f and m = −v/u. Distances are measured from the pole P. Concave mirror: f negative. Convex mirror: f positive.

Lens1/v − 1/u = 1/f and m = v/u. Distances are measured from the optical centre O. Convex lens: f positive. Concave lens: f negative.

Both use the same New Cartesian sign convention, and u is always negative. Mirror: plus in the formula, minus in m. Lens: minus in the formula, no minus in m.

Signs and power

uAlways negative (the object is on the left of O)

vPositive: real image on the far side. Negative: virtual image on the object's side

mNegative: real and inverted. Positive: virtual and erect

Power PP = 1/f, with f in metres. Power is how strongly a lens converges or diverges light: a shorter focal length means more power

Unit of powerDioptre (D). 1 D is the power of a lens whose focal length is 1 m; 1 D = 1 m⁻¹

Sign of powerConvex lens: positive. Concave lens: negative

Lenses in contactNet power P = P₁ + P₂ + P₃ + ..., added with their signs

Convex lens: a real, enlarged image

Problem

An object 2 cm tall is placed 25 cm from a convex lens of focal length 20 cm. Find the position, size and nature of the image, and the power of the lens.

Working

Given (with signs): h = +2 cm, u = −25 cm, f = +20 cm (convex).
Lens formula: 1/v − 1/u = 1/f, so 1/v = 1/f + 1/u.
Substitute: 1/v = 1/20 + 1/(−25) = 1/20 − 1/25.
LCM is 100: 1/v = (5 − 4)/100 = 1/100, so v = +100 cm.
m = v/u = 100/(−25) = −4; h′ = m × h = −4 × 2 cm = −8 cm.
Power: f = 20 cm = 0.20 m, so P = 1/0.20 = +5 D.

Answer

The image is 100 cm on the other side of the lens. It is real, inverted and enlarged: 8 cm tall, four times the object. The power of the lens is +5 D.

Concave lens: a virtual, smaller image

Problem

An object 3 cm tall is placed 20 cm from a concave lens of focal length 10 cm. Find the position, size and nature of the image, and the power of the lens.

Working

Given: h = +3 cm, u = −20 cm, f = −10 cm (concave).
1/v = 1/f + 1/u = −1/10 + 1/(−20) = −2/20 − 1/20 = −3/20.
So v = −20/3 cm ≈ −6.7 cm (rounded to one decimal place).
m = v/u = (−20/3) ÷ (−20) = +1/3; h′ = (1/3) × 3 cm = +1 cm.
Power: f = −10 cm = −0.10 m, so P = 1/(−0.10) = −10 D.

Answer

The image is about 6.7 cm from the lens, on the same side as the object. It is virtual, erect and one-third the size (1 cm tall). The power is −10 D.

Activity: find the focal length of a convex lens

Work near a window on a bright day. Use a far-off tree or building as the object.

1
Fix the convex lens upright in a holder on a table. Place a white screen (a card) behind it.
2
Point the lens at a distant object, such as a tree or a tall building seen through the window.
Never look at the Sun through a lens. A lens can focus sunlight enough to burn paper.
3
Move the screen slowly towards or away from the lens until a sharp, small, inverted image of the distant object appears.
4
Measure the distance between the centre of the lens and the screen with a metre scale.
5
Repeat three times and take the average. This is the approximate focal length of the lens.
Rays from a very distant object are almost parallel, so they meet at F₂.
6
Find the power: P = 1/f, with f in metres. For example, f = 25 cm = 0.25 m gives P = +4 D.

In real life

When an optician writes +2.0 D or −1.5 D on a spectacle prescription, that is the power of the lens: plus means convex, minus means concave. A −1.5 D lens has a focal length of about −0.67 m.

Exam angle

1 marks: “Two lenses of power +3.5 D and −1.5 D are placed in contact. What is the net power, and does the pair converge or diverge light?” P = +3.5 − 1.5 = +2.0 D. Positive, so the pair converges light, like a convex lens of focal length 0.5 m.

2 marks: “Define 1 dioptre. Find the power of a concave lens of focal length 25 cm.” 1 D is the power of a lens whose focal length is 1 m. f = −0.25 m, so P = 1/(−0.25) = −4 D.

3 marks: “An object is placed 20 cm from a concave lens of focal length 20 cm. Find the image distance and magnification, and state the nature of the image.” u = −20 cm, f = −20 cm; 1/v = −1/20 − 1/20 = −1/10; v = −10 cm (same side); m = v/u = +0.5: virtual, erect, half size.

Remember it

A lens has a line (minus) in its formula: 1/v − 1/u. A mirror adds up: 1/v + 1/u. Power needs metres: change cm to m before you find 1/f.